# Proof of Identity to Zero of the Sum of a Product of Binomial Coefficients & Pochhammer Numbers

It's well-know that the sum across an entire row of binomial coefficients (of degree, say, $$n$$) with alternating signs attached is 0; and it can easily be proven by demonstrating that it is the binomial expansion of $$(1-1)^n$$. What is less obvious is that if you take the same series & multiply the terms in it by consecutive Pochhammer numbers of degree $$n-1$$, but from anywhere along that series, the sum is still zero! To put it formally $$\sum_{k=0}^n \frac{(-1)^k(m+k)_{n-1}}{k!(n-k)!}=0 ,$$ $$∀m\geq n-1$$, it not mattering that the $$n!$$ is omitted in the denominator, as only the relative size of the terms matters, the sum being putatively 0. That this is so follows from a tracing of the combinatorial consequence on the terms in the series expansion about 1 of the logarithm of the property of the logarithm, that the logarithm of a product = the sum of the logarithms; so if we have a proof through some other route (which we do) that the property & the series imply each other, then we effectively already have a proof of the theorem adduced as the grounds of this post ... but an exceedingly roundabout proof! There must be a more elementary proof than that ... but I can't figure it out. I have a very vague recollection of seeing one when reading elementary stuff on binomial coefficients a long time ago; but I'm not absolutely certain I did - it might have been a proof of some other theorem.

It did occur to me that one route through which it might be proven that of recasting the $$n$$th row (the very top one with $$1$$ only in it being the zeroth) of the binomial coefficient triangle as being the series $$1+\sum_{k=0}^{n-1}\prod_{l=0}^k\frac{l-k}{l+1} ,$$ whence the one queried in this post being recast as $$1+\sum_{k=0}^{n-1}\prod_{l=0}^k\frac{(l-k)(l+m+n)}{(l+1)(l+m+1)} ∀m∊ℕ_0$$ the absolute value mattering not atall, as it is only the relative value of the terms that matters, the sum being putatively zero.

I don't know that this is the best way to the proof, but it did occur to me it might be viable. I wonder whether anyone would show a proof or a clue to one, whether by this means or otherwise.

There might possibly be a clue in the fact that the lastmentioned series could just as well be $$1+\sum_{k=0}^{n-1}\prod_{l=0}^k\frac{(l-k)(l+m+1)}{(l+1)(l+m+n)} ∀m∊ℕ_0$$ as because the binomial coefficients are symmetrical from front to back, it matters not in which direction the series of Pochhammer numbers is traversed.

• What exactly is the identity you're trying to prove? – Robert Israel Dec 3 '18 at 16:22
• The sum of alternating binomial coefficients (of degree say, n) is 0 - we know that, & it's easy to prove. But if you take a subsequence of consecutive Pochhammer numbers of degree n-1 (products of n-1 consecutive integers ... or polyhedral numbers of degree n-1, if you prefer, they're just Pochhammer numbers ÷by (n-1)!) and multiply each (alernating) binomial coefficient by each Pochhammer number in order, traversing each sequence one at a time, the sum is still zero. It''ll no doubt transpire to be elementary binomial coefficient stuff! – AmbretteOrrisey Dec 3 '18 at 16:40
• If you have a term for $l=k$ and one of the factors in the numerator is $l-k$, the product will be $0$. I don't think that's what you mean. – Robert Israel Dec 3 '18 at 16:42
• I've clarified what theorem I mean in the body of the question. ¶ Right yes! that going to zero when l=k, that just makes the sequence 0 thereafter ... as it ought to, as it's the binomial coefficients. Those expressions with the sum & product signs in are just a distillaton of the formula with the factorials & Pochammer symbols in - going to the next pochhamner number is equivalent to dividing by the smallest factor in the present one & multiplying by the largest in the next one ... or multipltying by 1>than the largest in the present one. – AmbretteOrrisey Dec 3 '18 at 17:02
• It's really frustrating actually, coz I can easily verify it with a spreadsheet, and I know it's not particularly advanced binomial-coefficient theory (in fact it's extremely elementary compared to how stratospheric some of that stuff gets!), but I just can't nail how to actually do it. In standard texts it might be a bit disguised relative to my formulation of it (in terms of Pochamner numbers) - you might well find it done purely in terms of binomial coefficients - but essentially the same theorem. – AmbretteOrrisey Dec 3 '18 at 17:39

Starting from

$$\sum_{k=0}^n (-1)^k \frac{1}{k!(n-k)!} (m+k)^\underline{n-1}$$

we get

$$\sum_{k=0}^n (-1)^k \frac{(n-1)!}{k!(n-k)!} {m+k\choose n-1} = \frac{1}{n} \sum_{k=0}^n (-1)^k {n\choose k} {m+k\choose n-1} \\ = \frac{1}{n} \sum_{k=0}^n (-1)^k {n\choose k} [z^{n-1}] (1+z)^{m+k} \\ = [z^{n-1}] (1+z)^{m} \frac{1}{n} \sum_{k=0}^n (-1)^k {n\choose k} (1+z)^k \\ = [z^{n-1}] (1+z)^{m} \frac{1}{n} (1-(1+z))^n \\ = \frac{(-1)^n}{n} [z^{n-1}] z^n (1+z)^{m} = 0.$$

Here we have used the fact that $$z^n (1+z)^{m} = z^n + \cdots$$ note however that we are extracting the coefficient on $$[z^{n-1}].$$ Careful inspection reveals that this will go through for $$n\ge 1$$ and all integer $$m$$ including negative. This is because $$(1+z)^m$$ is entire when $$m\ge 0$$ and the only pole of $$(1+z)^m$$ when $$m\lt 0$$ is at $$z=-1.$$

Observe that this is the combinatorial interpretation of the Pochhammer symbol as the falling factorial as opposed to the one from special functions. These two are documented at MathWorld. The proof for the alternate notation is left to the reader.

• That is slick that is! I'll need to cast it in my mind (as the King James Bible is very fond of saying!) to get it properly 'of a piece'. (Nearly said 'integrated' ... but that might have been confusing on a mathematics thread!) Sort of a generating function method. Like I said, I had effectively proven it by showing that it's a necessary condition for the logarithm to have its property given its Taylor series ... which is effectively another absurdly complicated quasi generator-function method ... but there just had to be something simpler! Thankyou for that. – AmbretteOrrisey Dec 3 '18 at 18:26

The forward Finite Difference of a function $$f(x)$$ is defined as $$\Delta \,f(x) = f(x + 1) - f(x)$$ and its iteration as $$\Delta ^{\,n} \,f(x) = \Delta \left( {\,\Delta ^{\,n - 1} \,f(x)} \right) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{n - k} \left( \matrix{ n \cr k \cr} \right)\;f(x + k)}$$

It is a known fact that if $$f(x)=p_q(x)$$ is a polynomial of degree $$q$$ then, similarly to derivatives, $$\Delta ^{\,n} \,p_{\,q} (x) = 0\quad \left| {\;q < n} \right.$$

So the case you propose is just \eqalign{ & \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^k \;} {{\left( {x + k} \right)_{\,n - 1} } \over {k!\left( {n - k} \right)!}} = \left( { - 1} \right)^n n!\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{n - k} \left( \matrix{ n \cr k \cr} \right)\;(x + k)_{n - 1} } _\, = \cr & = \left( { - 1} \right)^n n!\Delta ^{\,n} \,x^{\,\overline {\,n - 1\,} } = \left( { - 1} \right)^n n!\Delta ^{\,n} \prod\limits_{0\, \le \,j\, \le \,n - 2} {\left( {x + j} \right)} \, = 0 \cr} with x integer, real or complex.

• And that implies that the theorem infact holds for the degree of the Pochhammer numbers being less than n-1 also - that 1 & n-1 are not special cases. And ... I know you have used x as a convenience ... or are you? ... no, you're showing me that it's true also for general real values of x as well as naturals! Thankyou for that - it's a very thorough answer - it's well dissipated the fog! – AmbretteOrrisey Dec 3 '18 at 18:04
• @AmbretteOrrisey: glad to have helped and yes, $x$ can be even .. complex .. and even a matrix. The Delta (forward and backward) formula is quite basic to discrete maths: I do suggest you to get well acquainted with it. So.. you can discover by yourself what you get when taking the nth-Delta of a degree-n polynomial – G Cab Dec 3 '18 at 18:16
• It is something I've looked-at before ... but not enough really - never is quite enough! - and I've even used it often enough for computing derivatives numerically, & similar things. But I get too easily putoff sonetimes by the more rarified theory of things like that. Altough what you've just shown me is maturing & settling in my mind very nicely now. – AmbretteOrrisey Dec 3 '18 at 18:34
• I'm not too keen on this business of only being able to vote for one answer though ... so I'm just leaving that. I'll stick to saying what I think through comments. And that applies generally. ¶ Yes - I've seen how you can have e^∆, & stuff like that. – AmbretteOrrisey Dec 3 '18 at 18:38